Hi everyone! In this page https://en.wikipedia.org/wiki/Skolem_normal_form there is a part with "... by the axiom of choice ..." - Is AC even required? Can't we just say that, there is an element in the model which satisfies this formula, let's say b, then we create the constant b function

I'm trying to find the "geometric place" of all points of $C$ so the angle is $ACB=120$ using analytical geometry. I took three points on a circle. So If the angle $ACB$ is 120 degrees then $AOB$ is also 120 degrees. But now I'm stuck.

As I understand I need to set $C$ to be $(x_C,y_C)$ and somehow to get an equation using only those points.

I though of using the cos theorem, but I also have $A$ or $B$ in the final equation.

In geometry, an inscribed angle is the angle formed in the interior of a circle when two secant lines (or, in a degenerate case, when one secant line and one tangent line of that circle) intersect on the circle. It can also be defined as the angle subtended at a point on the circle by two given points on the circle.
Equivalently, an inscribed angle is defined by two chords of the circle sharing an endpoint.
The inscribed angle theorem relates the measure of an inscribed angle to that of the central angle subtending the same arc.
== Theorem ==
=== Statement ===
The inscribed angle theorem states...

Here is what I think is a short argument, no explicit numbers or painful fiddling

Let A be a finite abelian group. Let x be an element of order m. Let y be an element of maximal order n. If (m,n) = m, then we are done, as this means m divides n. If m > (m,n) > 1, then replacing x by (m,n)x, we see that there is an element z of order k with (k, n) = 1. I claim this is impossible.

For consider the element z+y. We have kn(z+y) = 0, so ord(z+y) | kn. If k' is a proper divisor of k, then k'n(z+y) = (k'n)z, which is not zero, as k'n is not zero in Z/k (here we use that (k,n) = 1 and that k' is a proper divisor). Similarly for any divisor n' of n. Thus ord(z+y) = kn > n; this contradicts the assumption that y had maximal order.

the proof I know uses the polynomial $p(t) = \prod\limits_{i=1}^r (\alpha+t\beta - \sigma_i(\alpha)- t\sigma_i(\beta))$, where $\sigma_i$ is an embedding of the extension into is algebraic closure and argues that since this polynomial has finitely may roots, there's a $c \in E(\alpha, \beta)$ which is not fixed by any of the embeddings

and I honestrly can't remember why, but this implies that $E(\alpha, \beta) = E(c)$. (maybe something using polynomials? is it where separability is needed?)

I think what's going on is that the image of an element $\alpha$ under an embedding $\sigma$ is a zero of the minimal polynomial of $\alpha$ still. So if no $\sigma$ fixes $\alpha$, the extension $K(\alpha)/K$ has degree at least the number of $\sigma$s. For $\alpha$ to be primitive, you want this to be the degree of your initial extension, say $L/K$. There are precisely enough $\sigma$s for this to work (namely $[L\colon K]$ many) if the extension is separable.

Also, mainly @Ted and @Thorgott: I apologize for being so insistent and bragging about the same topics over and over. I’m really interested in the topics I’m studying but my book really does not help me, and I don’t know better resource.

Suppose a function f is holomorphic in some disc, then if I write the power series of f about the center of this disc, will the radius of convergence of this power series be atleast as much as the radius of the disc?

supppose wlog that the center of the disk is $0$ to make notation less cumbersome. Let $C$ be circle around $0$ traversed once and of radius $<r$, where $f$ is holomorphic on the disk with radius $r$. Then we have the Cauchy integral formula $f(z)=\frac{1}{2\pi i}\int_C\frac{f(w)}{z-w}\mathrm{d}w$

Now the idea is the following: use that $\frac{1}{w-z}=\frac{1}{w} \frac{1}{1-\frac{z}{w}}=\frac{1}{w}(1+\frac{z}{w}+\frac{z^2}{w^2} + \dots+(\frac{z}{w})^{n-1}+\frac{(\frac{z}{w})^n}{1-\frac{z}{w}})$ (this is just a partial geometric series)

(which is obtained from just differentiating the usual Cauchy integral formula)

apply that with $z=0$ and compare with the terms we have in our integral, then we see that $f(z)=\sum_{k=0}^n \frac{f^{(k)}(0)z^k}{k!}+R$, where $R$ is the remainder term $\frac{1}{2\pi i} \int_C \frac{z^n}{w^n} \frac{f(w)}{w-z} \mathrm{d}w$

so all that is left to show is that $R \to 0$ as $n \to \infty$

I'll leave that to you :P

it's just some standard inequalities you have to deal with a lot in complex analysis

@LukasHeger I'm trying to understand the concept of a generic point corresponding to the zero ideal, as it pops up in the definition of a principal divisor.

@LukasHeger In Hartshorne, a principal divisor is defined as $(f) = \sum \nu_Y(f) Y$ where $\nu_Y$ is a discrete valuation for the discrete valuation ring $\mathcal O_{\eta,X}$ where $\eta$ is the generic point on $Y \subset X$.

I understand it in the case that the prime divisors are points, then the discrete valuation is just the order of the poles or zeroes. But in the general case, in practice, what do you take as the discrete valuation?

@JamalS if you have a regular point on a curve, then the stalk at that point is automatically a discrete valuation ring with quotient field $\mathcal{O}_{\eta,X}$ (which is the function field of the $Y$)

I think in general the answer to your question is just an algebraic coincidence: if $X$ is normal, then $\mathcal O{\eta,X}$ happens to be a discrete valuation ring, because its Krull dimension is $1$ and it is a normal domain and these happen to be discrete valuation rings

there is not necessarily a simple description of the discrete valuation in geometric terms

well, if you know the maximal ideal of your discrete valuation ring, then the discrete valuation is easy to describe: if $f \in \mathcal O_{\eta,X}$, then the normalized discrete valuation is the biggest integer $n$, such that $f$ is contained in the $n$-th power of the maximal ideal

all I'm saying is that there is no particular "geometric" description for the discrete valuation

@JamalS maybe it's helpful if you look at a somewhat non-geometric example, say $X=\mathrm{Spec}(\Bbb Z)$, pick a prime $p$, then we have a closed point $(p)$ in $X$ which is a prime divisor if we consider it as the closed subscheme $\mathrm{Spec}(\Bbb F_p)$

the stalk is the localization $\Bbb Z_(p)$ and the discrete valuation is the usual $p$-adic valuation

Now if you take a rational number, say $\frac{a}{b}$, then you can reasonably think of the $p$-adic valuation as counting poles and zeroes of that rational number as a "meromorphic function" on $X$

for example, $\frac{5}{6}$ has a zero of order $1$ at $(5)$ and poles at $(2)$ and $(3)$

at least we can think of it like that through analogy

we can think of $\frac{5}{6}$ as a "meromorphic function" on $X$, because if we have a point $(p) \in X$, then if $p$ doesn't divide $6$ (where this function has a pole), then $\frac{5}{6}$ makes sense in $\Bbb Z/p\Bbb Z$

so in this sense, the notions of poles and zeroes do kind of make sense in general

He explains what he calls the Thom-Smale-Witten complex, where given a Morse-Smale function $f : M \to \Bbb R$, you let $C_k^f(M)$ be the free abelian group generated by the index $k$ critical points and $\partial : C_k^f(M) \to C_{k-1}^f(M)$ is given by $\partial a = \sum_b \varepsilon_{ab} b$ where $b$ varies over index $k-1$ critical points and $\varepsilon_{ab}$ is the signed count of the number of instantons (flowlines of $\nabla f$) exiting $a$ and entering $b$.

hello. If $R$ is a ring, and $M$ is an $R$-module with generators $e_1,\dots,e_n$, and $M$ is free (and we ask in that case that the generators are linearly independent), then the map $f:R^n\to M$ given by $(r_1,\dots,r_n)\mapsto r_1e_1+\dots+r_ne_n$ is an isomorphism. If $M$ isn't free though, then that map is only surjective, but has some kernel, giving us $R^n/\text{ker}(f)\cong M$. Is this what is meant by giving generators and relations for $M$, i.e. that...cont

$M=<e_1,\dots,e_n>/<q_1,\dots,q_m>$ where $q_i = r_{1,i}e_1+\dots + r_{n,i}e_n$ for generators $(r_{1,i},\dots,r_{n,i})\in\text{ker}(f)$ of the kernel of $f$.

(I know how one can formulate this more normally, but I was specifically wanting to relate it to the quotient of a free module we get for finitely generated modules)

I dunno how to formally do the signed count though. I think it's like, for any critical point $c$ let $S_c$ and $U_c$ be the stable and unstable manifolds of the flow of $\nabla f$ at $c$. Since $f$ is Morse-Smale, $S_a$ and $U_b$ are always transverse for any pair $a, b$ of critical points. So $\dim (S_a \cap U_b) = \lambda_a - \lambda_b$, difference of the indices. If this difference is $1$ then we have finitely many flowlines from $a$ to $b$.

Now if $\gamma$ is a flowline from $a$ to $b$, then for any $x \in \gamma$, there is a short exact sequence $0 \to T_x \gamma \to T_x S_a \oplus T_x U_b \to T_x M \to 0$

The orientation on $T_x \gamma$ comes from the direction of the flow of $\nabla f$, and the same is true of $T_x S_a$ and $T_x U_b$. I guess I count $+1$ if the exact sequence preserves orientation, $-1$ otherwise

Because $S_a = \{x \in M : \lim_{t \to -\infty} \phi^t(x) = a\}$, so it's a union of a bunch of flowlines which "converge" to $a$, so doesn't those tangent directions to the flowlines give some orientation?

Each term in $\partial^2 a$ counts the total number of 2-broken flowlines from an index $k$ critical point to an index $k-2$ critical point. These should be zero somehow, because $M(a, b) \cup M(b, c)$ for a triple of critical points $a, b, c$ each having index less than the previous one is a manifold which can be naturally "compactified" to a manifold with boundary consisting of the 2-broken flowlines from $a$ to $b$ to $c$.

Once I quotient by the $\Bbb R$-action by flow, it's a 1-manifold with boundary.

That has an even number of boundary components, cancelling in pairs. Something of this sort.

For orientations, if $b$ and $a$ have index difference 1 and $\gamma$ is a flowline between them, we have as vector spaces $T\gamma = TU_b \cap TS_a$, right? So the sign should be comparing the orientation of these.

The first is oriented by direction. For the second we need a consistent rule to orient the intersection of transverse subspaces. There is some standard way, but I don't know what properties you want it to have; probably associativity, and I am too lazy to check if this way does that.

But let $V,W \subset T$ be transverse subspaces of an oriented vector space. Write $C$ for their intersection. Then we may write $V = C + V'$ and $W = C + W'$ for subspaces $V',W'$ which intersect $C$ trivially. We have $T = C + V' + W'$. There is a unique way to orient $C$, and thus $V',W'$, so that the above isomorphism with $T$ is oriented.

I don't know exactly what you want to do. If V, W are oriented subspaces of T then there is a short exact sequence $0 \to V \cap W \to V \oplus W \to T \to 0$. I oriented $V \cap W$ according to this rule.

That's the standard technique

I thought the problem here is we want a canonical orientation on the stable manifolds

(So $V$ and $W$ don't have canonical orientations given)

I wonder how one would build such a complex for a stratified space. I suppose a Morse function $f : W \to \Bbb R$ on a Whitney stratified space $W$ would be a stratumwise Morse function such that $df$ doesn't vanish on the limiting tangent planes on the lower strata.

I can have flowlines entering lower strata. Even worse, I might have critical points on different strata with equal index

I suppose $U_a$ and $S_a$ still make sense, they'll just be substratified spaces instead.

No, I cannot have flowlines entering lower strata, because if $A < B$ is a pair of strata, $d(f|_B)$ would not vanish on the limiting tangent planes to $A$. I can have flowlines converging to broken flowlines which enters lower strata, I suppose

The multivariable analysis course at UGA used to run very rarely. Once I started my course, all the potential undergraduate audience for that course had already had my course, and the one or two grad students who might have taken it weren't enough, obviously.